Because sinx\sin x is not a constant, and for what I am concerned with, it cannot be the coefficient of cosx\cos x. Therefore, for all the expressions of the type cos(nx+a)\cos(n x+ a) I am expecting to find zero, except when n=1n=1.

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This is not exactly what I think would work, I will explain it in the edit 🙂
– usumdelphini
Dec 8 ’15 at 10:31

@Kuba It would be zero.
– usumdelphini
Dec 8 ’15 at 10:46

@Kuba for all the expressions of the type cos(nx+a)\cos(nx+a) I am expecting to find zero, except when n=1n=1
– usumdelphini
Dec 8 ’15 at 10:49

These expressions do not have unique representations as polynomials in {sin(x),cos(x)}. This of course is due to the fundamental trig identity. A possible way around this would be to obtain coefficients in terms of both trig terms.
– Daniel Lichtblau
Dec 8 ’15 at 14:58

Something like findCoeff[Sin[2 x+a],Cos[2 x]] will still not work as you want it to, though. One probably has to fiddle with FourierCosCoefficient to do that.
In fact, I’m not even sure that the problem is well stated: consider for example Sin[2x]. How should it expand? FourierCosCoefficient[Sin[2 x],x ,1] gives the answer 8/3π8/3\pi, while findCoeff[Sin[2 x],Cos[x]] gives 0. While the former result answers a well defined question, what exactly are the rules according to which you want the latter result?